Engineering Examination (FE Exam) –
Practice Questions And Correct Answers
(Verified Answers) Plus Rationales 2026
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1. The derivative of f(x)=3x4−5x2+2f(x) = 3x^4 - 5x^2 + 2f(x)=3x4−5x2+2
is:
a) 12x3−10x12x^3 - 10x12x3−10x
b) 12x3−10x+212x^3 - 10x + 212x3−10x+2
c) 12x3−5x12x^3 - 5x12x3−5x
d) 3x3−5x3x^3 - 5x3x3−5x
Answer: a) 12x3−10x12x^3 - 10x12x3−10x
Rationale: The derivative of xnx^nxn is nxn−1nx^,n-1}nxn−1. So,
d/dx(3x4)=12x3d/dx(3x^4) = 12x^3d/dx(3x4)=12x3, d/dx(−5x2)=−10xd/dx(-
5x^2) = -10xd/dx(−5x2)=−10x, and d/dx(2)=0d/dx(2) = 0d/dx(2)=0.
, 2. A simply supported beam of length 6 m carries a uniformly distributed
load of 4 kN/m. The maximum bending moment occurs at:
a) The supports
b) Midspan
c) One-third span
d) Two-thirds span
Answer: b) Midspan
Rationale: For a simply supported beam with a uniform load, the maximum
bending moment occurs at midspan and equals wL2/8wL^2/8wL2/8.
3. If a fluid has a density of 1000 kg/m³ and a pressure of 200 kPa, the
pressure head is:
a) 20 m
b) 10 m
c) 15 m
d) 25 m
Answer: b) 20 m
Rationale: Pressure head h=P/(ρg)=200,000/(1000×9.81)≈20.39 mh =
P/(\rho g) = 200,000 / (1000 × 9.81) ≈ 20.39 \,
mh=P/(ρg)=200,000/(1000×9.81)≈20.39m.
, 4. The fundamental frequency of a simply supported beam is
proportional to:
a) L2L^2L2
b) 1/L21/L^21/L2
c) LLL
d) 1/L1/L1/L
Answer: b) 1/L21/L^21/L2
Rationale: For a beam, f∝EI/(ρAL4)f \propto \sqrt{EI/(ρAL^4)}f∝EI/(ρAL4),
simplifying to f∝1/L2f \propto 1/L^2f∝1/L2 for fixed E, I, ρ, and A.
5. A thermodynamic process occurs at constant pressure. This is called:
a) Isochoric
b) Isothermal
c) Isobaric
d) Adiabatic
Answer: c) Isobaric
Rationale: By definition, isobaric processes occur at constant pressure.
6. Which material property measures a material’s resistance to
deformation under tensile stress?
a) Ductility
b) Toughness
, c) Modulus of elasticity
d) Hardness
Answer: c) Modulus of elasticity
Rationale: Modulus of elasticity (Young’s modulus) quantifies stress/strain
relationship under tension.
7. The integral of ∫02(4x−1)dx\int_0^2 (4x - 1) dx∫02(4x−1)dx is:
a) 6
b) 8
c) 4
d) 2
Answer: a) 6
Rationale: ∫02(4x−1)dx=*2x2−x+02=(8−2)−0=6\int_0^2 (4x -1) dx = [2x^2 -
x]_0^2 = (8 -2) -0 = 6∫02(4x−1)dx=[2x2−x]02=(8−2)−0=6.
8. The Reynold’s number indicates:
a) Viscosity
b) Laminar or turbulent flow
c) Pressure drop
d) Flow rate